Problem: Solve for $z$, $ -\dfrac{z - 8}{20z^3} = -\dfrac{6}{16z^3} + \dfrac{8}{16z^3} $
Explanation: First we need to find a common denominator for all the expressions. This means finding the least common multiple of $20z^3$ $16z^3$ and $16z^3$ The common denominator is $80z^3$ To get $80z^3$ in the denominator of the first term, multiply it by $\frac{4}{4}$ $ -\dfrac{z - 8}{20z^3} \times \dfrac{4}{4} = -\dfrac{4z - 32}{80z^3} $ To get $80z^3$ in the denominator of the second term, multiply it by $\frac{5}{5}$ $ -\dfrac{6}{16z^3} \times \dfrac{5}{5} = -\dfrac{30}{80z^3} $ To get $80z^3$ in the denominator of the third term, multiply it by $\frac{5}{5}$ $ \dfrac{8}{16z^3} \times \dfrac{5}{5} = \dfrac{40}{80z^3} $ This give us: $ -\dfrac{4z - 32}{80z^3} = -\dfrac{30}{80z^3} + \dfrac{40}{80z^3} $ If we multiply both sides of the equation by $80z^3$ , we get: $ -4z + 32 = -30 + 40$ $ -4z + 32 = 10$ $ -4z = -22 $ $ z = \dfrac{11}{2}$